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Full text of "Handbook Of Chemical Engineering - I"

REFRACTORIES
475
and bring out the fundamental chemical characteristics. In the case of aluminum, silicates the aluminum oxide is usually made equal to unity in the formula and the fluxing oxides combined in one expression so that the entire chemical composition is represented by only two variables, the RO and RaO oxides, and the silica. With basic refractories the principal constituent is made equal to unity. Thus for magne-site the formula would be: xHzO-RQ-yHzOz-zSiOz, where the R20 is usually negligible. The calculation involved consists in dividing the per cent of each constituent by its molecular weight, adding the quotients or molecular equivalents of each group and dividing every quotient by the equivalent of the oxide or group of oxides to be taken as unity. In the case of fire clays the alumina equivalent is taken as unity. The result is then written in the form of the expression: zROR263- 2/RO2. The iron oxide is usually given in the ferrous form. This calculation is illustrated by the following example, dealing with the chemical composition of a New Jersey firebrick, the first column of figures showing the analysis, the last the quotient obtained by dividing the per cent of each compound by its molecular weight in round numbers:
Silica...................  77.82 per cent -T-   60 = 1.296
Alumina................  19.00 per cent -r- 102 = 0.186
Ferric oxide.............    1.01 per cent ^ 160 = 0.00625 X 2 = 0.0125
(ferrous oxide)
Titanium oxide..........    1.65 per cent -r-   80 = 0.0206
Lime....................    0.22 per cent    56 = 0.0039
Magnesia...............    0.06 per cent    40 = 0.0015
Soda....................    0.10 per cent    62 = 0.0016
Potash..................    0.28 per cent    94 = 0.0030
Dividing through by the equivalent of alumina, 0.186, we have:
Silica.............................   1.296     4-0.186 = 6.96700
Alumina.........................  0.186     -v- 0.186 = 1.00000
Ferrous oxide.....................  0.0125   * 0.186 = 0.06720
Titanium oxide...................  0.0206   4- 0.186 = 0.1107ft
Lime............................  0.0039   + 0.186 = 0.02096
Magnesia........................  0.0015   -r- 0.186 = 0.00806
Soda............................  0.0016   *- 0.186 = 0.00806
Potash...........................  0.00298 -r- 0.186 = 0.01602
Adding the values obtained for the five fluxing constituents, we have:
Ferrous oxide (FeO).................................  0.06720
Lime.............................................. 0.02096
Magnesia..........................................  0.00806
Soda...............................................  0.00806
Potash.............................................  0.01602
0.12030
Arranging the three groups in the empirical formula, we obtain:
6.97Si02
1A1203
0.11 Ti02
0.016 K20 0.008 Na2O 0.067 FeO 0.021 CaO 0.008 MgO
Reducing this expression to its simplest terms we obtain: 0.12 RO-AlsOs-6.97 Si02
1
It is evident that we are dealing here with a siliceous clay having an excess of 4.97 molecular equivalents of free silica over and above the combined silica of the